Difference between revisions of "1961 AHSME Problems/Problem 1"

(Solution to Problem 1)
Line 1: Line 1:
When simplified <math>(-\frac{1}{125})^{-\frac{2}{3}}</math> becomes:
+
== Problem 1==
 +
 
 +
When simplified, <math>(-\frac{1}{125})^{-2/3}</math> becomes:
 +
 
 +
<math>\textbf{(A)}\ \frac{1}{25} \qquad
 +
\textbf{(B)}\ -\frac{1}{25} \qquad
 +
\textbf{(C)}\ 25\qquad
 +
\textbf{(D)}\ -25\qquad
 +
\textbf{(E)}\ 25\sqrt{-1}</math> 
 +
 
 +
==Solution==
 +
To remove the negative exponent, flip the fraction of the base.  This results in <math>(-125)^{2/3}</math>.
 +
 
 +
Then, cube root <math>-125</math> and and square the result to get the answer of <math>25</math>, or answer choice <math>\boxed{\textbf{(C)}}</math>.
 +
 
 +
==See Also==
 +
{{AHSME 40p box|year=1961|before=First Question|num-a=2}}
 +
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:04, 17 May 2018

Problem 1

When simplified, $(-\frac{1}{125})^{-2/3}$ becomes:

$\textbf{(A)}\ \frac{1}{25} \qquad \textbf{(B)}\ -\frac{1}{25} \qquad \textbf{(C)}\ 25\qquad \textbf{(D)}\ -25\qquad \textbf{(E)}\ 25\sqrt{-1}$

Solution

To remove the negative exponent, flip the fraction of the base. This results in $(-125)^{2/3}$.

Then, cube root $-125$ and and square the result to get the answer of $25$, or answer choice $\boxed{\textbf{(C)}}$.

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
All AHSME Problems and Solutions


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS